Derivative of a function f at a number a
edit
We denote the derivative of a function
f
{\displaystyle f}
at a number
a
{\displaystyle a}
as
f
′
(
a
)
{\displaystyle f'(a)\,\!}
.
The derivative of a function
f
{\displaystyle f}
at a number
a
{\displaystyle a}
a is given by the following limit (if it exists):
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
{\displaystyle f'(a)=\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}}
An analagous equation can be defined by letting
x
=
(
a
+
h
)
{\displaystyle x=(a+h)}
. Then
h
=
(
x
−
a
)
{\displaystyle h=(x-a)}
, which shows that when
x
{\displaystyle x}
approaches
a
{\displaystyle a}
,
h
{\displaystyle h}
approaches
0
{\displaystyle 0}
:
f
′
(
a
)
=
lim
x
→
a
f
(
x
)
−
f
(
a
)
x
−
a
{\displaystyle f'(a)=\lim _{x\rightarrow a}{\frac {f(x)-f(a)}{x-a}}}
As the slope of a tangent line
edit
Given a function
y
=
f
(
x
)
{\displaystyle y=f(x)\,\!}
, the derivative
f
′
(
a
)
{\displaystyle f'(a)\,\!}
can be understood as the slope of the tangent line to
f
(
x
)
{\displaystyle f(x)}
at
x
=
a
{\displaystyle x=a}
:
Find the equation of the tangent line to
y
=
x
2
{\displaystyle y=x^{2}}
at
x
=
1
{\displaystyle x=1}
.
To find the slope of the tangent, we let
y
=
f
(
x
)
{\displaystyle y=f(x)}
and use our first definition:
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
⇒
lim
h
→
0
(
a
+
h
)
2
−
(
a
)
2
h
⇒
lim
h
→
0
a
2
+
2
a
h
+
h
2
−
a
2
h
⇒
lim
h
→
0
h
(
2
a
+
h
)
h
⇒
lim
h
→
0
(
2
a
+
h
)
{\displaystyle f'(a)=\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}(a+h)^{2}-(a)^{2}}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}a^{2}+2ah+h^{2}-a^{2}}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}h(2a+h)}{h}}\Rightarrow \lim _{h\rightarrow 0}{\color {Blue}(2a+h)}}
It can be seen that as
h
{\displaystyle h}
approaches
0
{\displaystyle 0}
, we are left with
f
′
(
a
)
=
2
a
{\displaystyle f'(a)={\color {Blue}2a}\,\!}
. If we plug in
1
{\displaystyle 1}
for
a
{\displaystyle a}
:
f
′
(
1
)
=
2
(
1
)
⇒
2
{\displaystyle f'({\color {Red}1})=2({\color {Red}1})\Rightarrow {\color {Red}2}}
So our preliminary equation for the tangent line is
y
=
2
x
+
b
{\displaystyle y={\color {Red}2}x+b}
. By plugging in our tangent point
(
1
,
1
)
{\displaystyle (1,1)}
to find
b
{\displaystyle b}
, we can arrive at our final equation:
1
=
2
(
1
)
+
b
⇒
−
1
=
b
{\displaystyle {\color {Red}1}=2({\color {Red}1})+b\Rightarrow -1=b}
So our final equation is
y
=
2
x
−
1
{\displaystyle y=2x-1\,\!}
.
The derivative of a function
f
(
x
)
{\displaystyle f(x)}
at a number
a
{\displaystyle a}
can be understood as the instantaneous rate of change of
f
(
x
)
{\displaystyle f(x)}
when
x
=
a
{\displaystyle x=a}
.
The derivative as a function
edit
Consider the sequences:
y
=
x
2
;
y
+
Δ
y
=
(
x
+
h
)
2
=
x
2
+
2
x
h
+
h
2
{\displaystyle y=x^{2};\ y+\Delta y=(x+h)^{2}=x^{2}+2xh+h^{2}}
y
=
x
3
;
y
+
Δ
y
=
(
x
+
h
)
3
=
x
3
+
3
x
2
h
+
3
x
h
2
+
h
3
{\displaystyle y=x^{3};\ y+\Delta y=(x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}}
y
=
x
4
;
y
+
Δ
y
=
(
x
+
h
)
4
=
x
4
+
4
x
3
h
+
6
x
2
h
2
+
4
x
h
3
+
h
4
{\displaystyle y=x^{4};\ y+\Delta y=(x+h)^{4}=x^{4}+4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}}
y
=
x
5
;
y
+
Δ
y
=
(
x
+
h
)
5
=
x
5
+
5
x
4
h
+
10
x
3
h
2
+
10
x
2
h
3
+
5
x
h
4
+
h
5
{\displaystyle y=x^{5};\ y+\Delta y=(x+h)^{5}=x^{5}+5x^{4}h+10x^{3}h^{2}+10x^{2}h^{3}+5xh^{4}+h^{5}}
y
=
x
6
;
y
+
Δ
y
=
(
x
+
h
)
6
=
x
6
+
6
x
5
h
+
{\displaystyle y=x^{6};\ y+\Delta y=(x+h)^{6}=x^{6}+6x^{5}h\ +}
many terms containing
h
2
{\displaystyle h^{2}}
y
=
x
n
;
y
+
Δ
y
=
(
x
+
Δ
x
)
n
=
x
n
+
n
x
n
−
1
Δ
x
+
{\displaystyle y=x^{n};\ y+\Delta y=(x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x\ +}
many terms containing
(
Δ
x
)
2
{\displaystyle (\Delta x)^{2}}
Δ
y
=
n
x
n
−
1
Δ
x
+
{\displaystyle \Delta y=nx^{n-1}\Delta x\ +}
many terms containing
(
Δ
x
)
2
{\displaystyle (\Delta x)^{2}}
Therefore:
Δ
y
Δ
x
=
n
x
n
−
1
+
{\displaystyle {\frac {\Delta y}{\Delta x}}=nx^{n-1}\ +}
many terms with each and every term containing
Δ
x
{\displaystyle \Delta x}
d
y
d
x
=
lim
Δ
x
→
0
Δ
y
Δ
x
{\displaystyle {\frac {dy}{dx}}=\lim _{\Delta x\rightarrow 0}{\frac {\Delta y}{\Delta x}}}
d
(
x
n
)
d
x
=
n
x
n
−
1
{\displaystyle {\frac {d(x^{n})}{dx}}=nx^{n-1}}
read as "derivative with respect to
x
{\displaystyle x}
of
x
{\displaystyle x}
to the power
n
{\displaystyle n}
."
Later it will be shown that this is valid for all real
n
{\displaystyle n}
, positive or negative, integer or fraction.
Without Using Calculus
edit
Derivative of cubic function
edit
Graph of cubic function illustrating use of associated quadratic. When
x
==
7.5
,
{\displaystyle x==7.5,}
there is exactly 1 value of
x
{\displaystyle x}
that gives
f
(
x
)
=
f
(
7.5
)
.
{\displaystyle f(x)=f(7.5).}
When
x
==
3
,
{\displaystyle x==3,}
there are 3 values of
x
{\displaystyle x}
that give
f
(
x
)
=
f
(
3
)
.
{\displaystyle f(x)=f(3).}
When
x
==
p
,
{\displaystyle x==p,}
there are exactly 2 values of
x
{\displaystyle x}
that give
f
(
x
)
=
f
(
p
)
.
{\displaystyle f(x)=f(p).}
Point
(
p
,
f
(
p
)
)
{\displaystyle (p,f(p))}
is a stationary point.
In the diagram there is a stationary point at
x
=
p
.
{\displaystyle x=p.}
When
x
=
p
,
{\displaystyle x=p,}
there are exactly 2 values of
x
{\displaystyle x}
that produce
f
(
x
)
=
f
(
p
)
.
{\displaystyle f(x)=f(p).}
Aim of this section is to derive the condition that produces exactly 2 values of
x
.
{\displaystyle x.}
See Cubic function as
product
of linear function and quadratic.
The associated quadratic when
x
=
p
:
{\displaystyle x=p:}
A
=
a
{\displaystyle A=a}
B
=
A
p
+
b
=
a
p
+
b
{\displaystyle B=Ap+b=ap+b}
C
=
B
p
+
c
=
a
p
p
+
b
p
+
c
.
{\displaystyle C=Bp+c=app+bp+c.}
Divide
g
(
x
)
=
a
x
2
+
(
a
p
+
b
)
x
+
(
a
p
2
+
b
p
+
c
)
{\displaystyle g(x)=ax^{2}+(ap+b)x+(ap^{2}+bp+c)}
by
(
x
−
p
)
.
{\displaystyle (x-p).}
This division gives a quotient of
a
x
+
(
2
a
p
+
b
)
{\displaystyle ax+(2ap+b)}
and remainder of
h
(
p
)
=
3
a
p
2
+
2
b
p
+
c
.
{\displaystyle h(p)=3ap^{2}+2bp+c.}
Factor
(
x
−
p
)
{\displaystyle (x-p)}
divides
g
(
x
)
{\displaystyle g(x)}
exactly. Therefore, remainder
h
(
p
)
=
0
,
{\displaystyle h(p)=0,}
the desired condition.
When
x
=
p
,
h
(
x
)
=
3
a
x
2
+
2
b
x
+
c
,
{\displaystyle x=p,\ h(x)=3ax^{2}+2bx+c,}
the derivative of
f
(
x
)
.
{\displaystyle f(x).}
Derivative of quartic function
edit
Used where
y
=
U
(
x
)
V
(
x
)
{\displaystyle y={\frac {U(x)}{V(x)}}}
y
=
u
v
{\displaystyle y={\frac {u}{v}}}
y
v
=
u
{\displaystyle yv=u}
y
v
′
+
v
y
′
=
u
′
{\displaystyle yv'+vy'=u'}
v
y
′
=
u
′
−
y
v
′
=
u
′
−
u
v
′
v
{\displaystyle vy'=u'-yv'=u'-{\frac {uv'}{v}}}
y
′
=
u
′
v
−
u
v
′
v
2
=
v
u
′
v
2
−
u
v
′
v
2
=
v
u
′
−
u
v
′
v
2
{\displaystyle y'={\frac {u'}{v}}-{\frac {uv'}{v^{2}}}={\frac {vu'}{v^{2}}}-{\frac {uv'}{v^{2}}}={\frac {vu'-uv'}{v^{2}}}}
Derivatives of trigonometric functions
edit
y
=
sin
(
x
)
{\displaystyle y=\sin(x)}
y
+
Δ
y
=
sin
(
x
+
Δ
x
)
=
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
{\displaystyle y+\Delta y=\sin(x+\Delta x)=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)}
Δ
y
=
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
−
sin
(
x
)
{\displaystyle \Delta y=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)-\sin(x)}
Δ
y
Δ
x
=
sin
(
x
)
{\displaystyle {\frac {\Delta y}{\Delta x}}=\sin(x)}
⋅
cos
(
Δ
x
)
−
1
Δ
x
+
cos
(
x
)
{\displaystyle {\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)}
⋅
sin
(
Δ
x
)
Δ
x
{\displaystyle {\frac {\sin(\Delta x)}{\Delta x}}}
The value
cos
(
Δ
x
)
−
1
Δ
x
{\displaystyle {\frac {\cos(\Delta x)-1}{\Delta x}}}
:
>>> # python code
>>> [ ( math . cos ( Δx ) - 1 ) / Δx for Δx in ( .1 , .01 , .001 , .0001 , .0000_1 , .0000_01 , .0000_001 , .0000_0001 , .0000_0000_1 ) ]
[ - 0.049958347219742905 , - 0.004999958333473664 , - 0.0004999999583255033 ,
- 4.999999969612645e-05 , - 5.000000413701855e-06 , - 5.000444502911705e-07 ,
- 4.9960036108132044e-08 , 0.0 , 0.0 ]
>>>
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
=
lim
Δ
x
→
0
−
sin
(
Δ
x
)
1
=
0
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {\cos(\Delta x)-1}{\Delta x}}=\lim _{\Delta x\rightarrow 0}{\frac {-\sin(\Delta x)}{1}}=0}
by L'Hôpital's rule.
The value
sin
(
Δ
x
)
Δ
x
{\displaystyle {\frac {\sin(\Delta x)}{\Delta x}}}
:
>>> # python code
>>> [ math . sin ( Δx ) / Δx for Δx in ( .1 , .01 , .001 , .0001 , .0000_1 , .0000_01 , .0000_001 , .0000_0001 , .0000_0000_1 ) ]
[ 0.9983341664682815 , 0.9999833334166665 , 0.9999998333333416 ,
0.9999999983333334 , 0.9999999999833332 , 0.9999999999998334 ,
0.9999999999999983 , 1.0 , 1.0 ]
>>>
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
=
lim
Δ
x
→
0
cos
(
Δ
x
)
1
=
1
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {\sin(\Delta x)}{\Delta x}}=\lim _{\Delta x\rightarrow 0}{\frac {\cos(\Delta x)}{1}}=1}
by L'Hôpital's rule.
d
y
d
x
=
sin
(
x
)
{\displaystyle {\frac {dy}{dx}}=\sin(x)}
⋅
0
+
cos
(
x
)
{\displaystyle 0+\cos(x)}
⋅
1
=
cos
(
x
)
{\displaystyle 1=\cos(x)}
y
=
cos
(
x
)
=
1
−
sin
2
(
x
)
{\displaystyle y=\cos(x)={\sqrt {1-\sin ^{2}(x)}}}
y
2
=
1
−
sin
(
x
)
{\displaystyle y^{2}=1-\sin(x)}
⋅
sin
(
x
)
{\displaystyle \sin(x)}
Differentiate both sides:
2
y
{\displaystyle 2y}
⋅
d
y
d
x
=
−
(
2
sin
(
x
)
cos
(
x
)
)
{\displaystyle {\frac {dy}{dx}}=-(2\sin(x)\cos(x))}
d
y
d
x
=
−
sin
(
x
)
cos
(
x
)
cos
(
x
)
=
−
sin
(
x
)
.
{\displaystyle {\frac {dy}{dx}}=-{\frac {\sin(x)\cos(x)}{\cos(x)}}=-\sin(x).}
y
=
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle y=\tan(x)={\frac {\sin(x)}{\cos(x)}}}
y
cos
(
x
)
=
sin
(
x
)
{\displaystyle y\cos(x)=\sin(x)}
Differentiate both sides:
y
(
−
sin
(
x
)
)
+
cos
(
x
)
{\displaystyle y(-\sin(x))+\cos(x)}
⋅
d
y
d
x
=
cos
(
x
)
{\displaystyle {\frac {dy}{dx}}=\cos(x)}
cos
(
x
)
{\displaystyle \cos(x)}
⋅
d
y
d
x
=
cos
(
x
)
+
y
sin
(
x
)
{\displaystyle {\frac {dy}{dx}}=\cos(x)+y\sin(x)}
d
y
d
x
=
cos
(
x
)
+
tan
(
x
)
sin
(
x
)
cos
(
x
)
=
1
+
tan
2
(
x
)
=
sec
2
(
x
)
{\displaystyle {\frac {dy}{dx}}={\frac {\cos(x)+\tan(x)\sin(x)}{\cos(x)}}=1+\tan ^{2}(x)=\sec ^{2}(x)}
Derivatives of inverse trigonometric functions
edit
Derivatives of logarithmic functions
edit
a^x or
a
x
{\displaystyle a^{x}}
edit
y
=
a
x
{\displaystyle y=a^{x}}
y
+
Δ
y
=
a
x
+
Δ
x
=
a
x
a
Δ
x
{\displaystyle y+\Delta y=a^{x+\Delta x}=a^{x}a^{\Delta x}}
Δ
y
=
a
x
a
Δ
x
−
a
x
=
a
x
(
a
Δ
x
−
1
)
{\displaystyle \Delta y=a^{x}a^{\Delta x}-a^{x}=a^{x}(a^{\Delta x}-1)}
Δ
y
Δ
x
=
a
x
⋅
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}=a^{x}\cdot {\frac {a^{\Delta x}-1}{\Delta x}}}
Consider the value
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {a^{\Delta x}-1}{\Delta x}}}
specifically
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}}
. L'Hôpital's rule cannot be used here because
d
(
a
x
)
d
x
{\displaystyle {\frac {d(a^{x})}{dx}}}
is what we are trying to find.
a
1
2
=
a
{\displaystyle a^{\frac {1}{2}}={\sqrt {a}}}
(
a
1
2
)
1
2
=
a
{\displaystyle (a^{\frac {1}{2}})^{\frac {1}{2}}={\sqrt {\sqrt {a}}}}
a
1
2
3
=
a
{\displaystyle a^{\frac {1}{2^{3}}}={\sqrt {\sqrt {\sqrt {a}}}}}
a
1
2
n
=
a
{\displaystyle a^{\frac {1}{2^{n}}}={\sqrt {a}}}
taken
n
{\displaystyle n}
times.
We will look at the expression
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {a^{\Delta x}-1}{\Delta x}}}
for different values of
a
{\displaystyle a}
with
Δ
x
=
1
2
70
=
8.5
e
−
22
{\displaystyle \Delta x={\frac {1}{2^{70}}}=8.5e-22}
(approx) in which case the expression becomes
(
a
1
/
2
70
−
1
)
∗
(
2
70
)
{\displaystyle (a^{1/2^{70}}-1)*(2^{70})}
where
a
1
/
2
70
{\displaystyle a^{1/2^{70}}}
is
a
70
{\displaystyle {\sqrt[{70}]{a}}}
or
a
{\displaystyle {\sqrt {a}}}
taken
70
{\displaystyle 70}
times. This approach is used here because function sqrt()
can be written so
that it does not depend on logarithmic or exponential operations.
>>> # python code
>>> N = Decimal ( 2 )
>>> v2 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v2
Decimal ( '0.69314718055994530941743560122437474084363865015406919942144' )
>>>
>>> N = Decimal ( 8 )
>>> v8 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v8
Decimal ( '2.07944154167983592825352768227031325913255072732801782513664' )
>>>
>>> N = Decimal ( 32 )
>>> v32 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v32
Decimal ( '3.46573590279972654709124760144583715956091114572543812435968' )
>>>
>>> N = Decimal ( 128 )
>>> v128 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v128
Decimal ( '4.85203026391961716593059535875094644210510807293198187102208' )
>>>
Compare the values v8, v32, v128
with v2
:
>>> v8 / v2 ; v32 / v2 ; v128 / v2
Decimal ( '3.00000000000000000000176135549769209744528640235368520610520' )
Decimal ( '5.00000000000000000000587118499230699148996545343403093128046' )
Decimal ( '7.00000000000000000001232948848384468209997247971055570994695' )
>>>
We know that
8
=
2
3
;
32
=
2
5
;
128
=
2
7
{\displaystyle 8=2^{3};\ 32=2^{5};\ 128=2^{7}}
. The values v2, v8, v32, v128
are behaving like logarithms.
In fact
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}}
is the natural logarithm of
a
{\displaystyle a}
written as
ln
(
a
)
.
{\displaystyle \ln(a).}
Figure 5: Graphs of
a
x
−
1
x
{\displaystyle {\frac {a^{x}-1}{x}}}
for a =
2
,
8
,
32
,
128
{\displaystyle 2,\ 8,\ 32,\ 128}
.
Figure 5 contains graphs of
a
x
−
1
x
{\displaystyle {\frac {a^{x}-1}{x}}}
for
a
=
2
,
8
,
32
,
128
{\displaystyle a=2,\ 8,\ 32,\ 128}
with
graph of
e
x
−
1
x
{\displaystyle {\frac {e^{x}-1}{x}}}
included for reference.
All values of
x
{\displaystyle x}
are valid for all curves except where
x
=
0.
{\displaystyle x=0.}
The correct value of
ln
(
2
)
{\displaystyle \ln(2)}
is:
>>> Decimal ( 2 ) . ln ()
Decimal ( '0.693147180559945309417232121458176568075500134360255254120680' )
>>>
Our calculation produced:
Decimal ( '0.69314718055994530941743560122437474084363865015406919942144' )
accurate to 21 places of decimals, not bad for one line of simple python code using high-school math.
This method for calculation of
ln
(
a
)
{\displaystyle \ln(a)}
supposes that function sqrt()
is available.
Programming language python interprets expression a**b
as
a
b
.
{\displaystyle a^{b}.}
Therefore, in python,
ln
(
2
)
{\displaystyle \ln(2)}
can be calculated in accordance with the basic definition above:
# python code
>>> getcontext () . prec
101 # Precision of 101.
>>> dx = Decimal ( '1E-50' )
>>> ( 2 ** dx - 1 ) / dx
Decimal ( '0.69314718055994530941723212145817656807550013436026' ) # ln(2)
>>> ( 2 ** ( - dx ) - 1 ) / ( - dx )
Decimal ( '0.693147180559945309417232121458176568075500134360253' ) # ln(2)
>>>
When
y
=
a
x
,
d
y
d
x
=
a
x
⋅
ln
(
a
)
.
{\displaystyle y=a^{x},{\frac {dy}{dx}}=a^{x}\cdot \ln(a).}
The base of natural logarithms is the value of
a
{\displaystyle a}
that gives
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
=
1.
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}=1.}
This value of
a
{\displaystyle a}
, usually called
e
,
=
2.718281828459045235360287471352662497757247093699959574.....
{\displaystyle e,=2.718281828459045235360287471352662497757247093699959574.....}
>>> # python code
>>> N = e = Decimal ( 1 ) . exp (); N
Decimal ( '2.71828182845904523536028747135266249775724709369995957496697' )
>>> ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 )
Decimal ( '1.0000_0000_0000_0000_0000_0423_51647362715016770651530003719651328' )
>>> # ln(e) = 1. Our calculation of ln(e) is accurate to 21 places of decimals.
When
a
=
e
,
d
(
e
x
)
d
x
=
e
x
{\displaystyle a=e,{\frac {d(e^{x})}{dx}}=e^{x}}
⋅
ln
(
e
)
=
e
x
{\displaystyle \ln(e)=e^{x}}
⋅
1
=
e
x
.
{\displaystyle 1=e^{x}.}
y
=
l
n
(
x
)
{\displaystyle y=ln(x)}
x
=
e
y
{\displaystyle x=e^{y}}
d
x
d
y
=
e
y
{\displaystyle {\frac {dx}{dy}}=e^{y}}
d
y
d
x
=
1
e
y
=
1
x
{\displaystyle {\frac {dy}{dx}}={\frac {1}{e^{y}}}={\frac {1}{x}}}
y
=
ln
(
a
x
)
{\displaystyle y=\ln(ax)}
d
y
d
x
=
d
d
x
ln
(
a
x
)
=
d
d
x
(
ln
(
a
)
+
ln
(
x
)
)
=
d
d
x
ln
(
a
)
+
d
d
x
ln
(
x
)
=
1
x
{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\ln(ax)={\frac {d}{dx}}(\ln(a)+\ln(x))={\frac {d}{dx}}\ln(a)+{\frac {d}{dx}}\ln(x)={\frac {1}{x}}}
y
=
ln
(
x
a
)
{\displaystyle y=\ln(x^{a})}
d
y
d
x
=
d
d
x
ln
(
x
a
)
=
d
d
x
(
a
⋅
ln
(
x
)
)
=
a
⋅
d
d
x
ln
(
x
)
=
a
x
{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\ln(x^{a})={\frac {d}{dx}}(a\cdot \ln(x))=a\cdot {\frac {d}{dx}}\ln(x)={\frac {a}{x}}}
y
=
x
m
n
{\displaystyle y=x^{\frac {m}{n}}}
Calculate
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
y
n
=
x
m
{\displaystyle y^{n}=x^{m}}
n
⋅
ln
(
y
)
=
m
⋅
ln
(
x
)
{\displaystyle n\cdot \ln(y)=m\cdot \ln(x)}
n
⋅
1
y
⋅
d
y
d
x
=
m
⋅
1
x
{\displaystyle n\cdot {\frac {1}{y}}\cdot {\frac {dy}{dx}}=m\cdot {\frac {1}{x}}}
d
y
d
x
=
m
⋅
1
x
⋅
y
n
{\displaystyle {\frac {dy}{dx}}=m\cdot {\frac {1}{x}}\cdot {\frac {y}{n}}}
=
m
n
⋅
x
m
n
x
{\displaystyle ={\frac {m}{n}}\cdot {\frac {x^{\frac {m}{n}}}{x}}}
=
m
n
⋅
x
m
n
−
1
{\displaystyle ={\frac {m}{n}}\cdot x^{{\frac {m}{n}}-1}}
Careful manipulation of logarithms converts exponents into simple constants.
Used where
y
=
G
(
H
(
I
(
x
)
)
)
{\displaystyle y=G(H(I(x)))}
y
=
cos
(
2
x
)
{\displaystyle y=\cos(2x)}
Let
y
=
cos
(
u
)
{\displaystyle y=\cos(u)}
where
u
=
2
x
{\displaystyle u=2x}
.
d
y
d
x
=
d
y
d
u
⋅
d
u
d
x
=
−
sin
(
u
)
⋅
2
=
−
2
sin
(
2
x
)
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}=&{\frac {dy}{du}}\cdot {\frac {du}{dx}}\\=&-\sin(u)\cdot 2\\=&-2\sin(2x)\\\end{aligned}}}
y
=
sin
2
(
ln
(
5
x
)
)
{\displaystyle y=\sin ^{2}(\ln(5x))}
Let
y
=
u
2
{\displaystyle y=u^{2}}
where
u
=
sin
(
v
)
;
v
=
ln
(
w
)
;
w
=
5
x
{\displaystyle u=\sin(v);\ v=\ln(w);\ w=5x}
.
d
y
d
x
=
d
y
d
u
⋅
d
u
d
v
⋅
d
v
d
w
⋅
d
w
d
x
=
2
u
⋅
cos
(
v
)
⋅
1
w
⋅
5
=
2
sin
(
v
)
⋅
cos
(
ln
(
w
)
)
⋅
1
5
x
⋅
5
=
2
sin
(
ln
(
w
)
)
⋅
cos
(
ln
(
5
x
)
)
⋅
1
x
=
2
sin
(
ln
(
5
x
)
)
⋅
cos
(
ln
(
5
x
)
)
⋅
1
x
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}=&{\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dw}}\cdot {\frac {dw}{dx}}\\=&2u\cdot \cos(v)\cdot {\frac {1}{w}}\cdot 5\\=&2\sin(v)\cdot \cos(\ln(w))\cdot {\frac {1}{5x}}\cdot 5\\=&2\sin(\ln(w))\cdot \cos(\ln(5x))\cdot {\frac {1}{x}}\\=&2\sin(\ln(5x))\cdot \cos(\ln(5x))\cdot {\frac {1}{x}}\\\end{aligned}}}
Applications of the Derivative
edit
The first derivative of
f
(
x
)
=
y
′
{\displaystyle f(x)=y'}
or
f
′
(
x
)
.
{\displaystyle f'(x).}
As shown above,
f
′
(
x
)
{\displaystyle f'(x)}
at any point
x
1
{\displaystyle x_{1}}
gives the slope of
f
(
x
)
{\displaystyle f(x)}
at point
x
1
.
{\displaystyle x_{1}.}
f
′
(
x
1
)
{\displaystyle f'(x_{1})}
is the slope of
f
(
x
)
{\displaystyle f(x)}
when
x
=
x
1
.
{\displaystyle x=x_{1}.}
Figure 1: Diagram illustrating relationship between
f
(
x
)
=
x
2
−
x
−
2
{\displaystyle f(x)=x^{2}-x-2}
and
f
′
(
x
)
=
2
x
−
1.
{\displaystyle f'(x)=2x-1.}
When
x
=
0.5
,
{\displaystyle x=0.5,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
x
=
0
,
{\displaystyle x=0,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
−
1.
{\displaystyle f(x)=-1.}
When
x
=
1
,
{\displaystyle x=1,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
1.
{\displaystyle f(x)=1.}
Point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is absolute minimum.
In the example to the right,
y
=
f
(
x
)
=
x
2
−
x
−
2
{\displaystyle y=f(x)=x^{2}-x-2}
and
y
′
=
f
′
(
x
)
=
2
x
−
1.
{\displaystyle y'=f'(x)=2x-1.}
Of special interest is the point at which
f
′
(
x
)
{\displaystyle f'(x)}
or slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
f
′
(
x
)
=
0
,
x
=
0.5
{\displaystyle f'(x)=0,\ x=0.5}
and
f
(
x
)
=
0.5
2
−
0.5
−
2
=
−
2.25.
{\displaystyle f(x)=0.5^{2}-0.5-2=-2.25.}
The point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is called a critical point or stationary point of
f
(
x
)
.
{\displaystyle f(x).}
Because
y
′
{\displaystyle y'}
has exactly one solution for
y
′
=
0
,
f
(
x
)
{\displaystyle y'=0,\ f(x)}
has exactly one critical point.
The value of
y
{\displaystyle y}
at point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is less than
y
{\displaystyle y}
at both
(
0
,
−
2
)
,
(
1
,
−
2
)
.
{\displaystyle (0,-2),\ (1,-2).}
Therefore the critical point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is a minimum of
f
(
x
)
.
{\displaystyle f(x).}
In this curve
y
=
x
2
−
x
−
2
,
{\displaystyle y=x^{2}-x-2,}
the point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is both local minimum and absolute minimum.
Figure 1: Diagram illustrating relationship between
f
(
x
)
=
2
x
3
+
3
x
2
−
12
x
−
8
8
{\displaystyle f(x)={\frac {2x^{3}+3x^{2}-12x-8}{8}}}
and
f
′
(
x
)
=
6
x
2
+
6
x
−
12
8
.
{\displaystyle f'(x)={\frac {6x^{2}+6x-12}{8}}.}
When
x
=
−
2
{\displaystyle x=-2}
or
x
=
1
,
{\displaystyle x=1,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
x
=
−
3
,
{\displaystyle x=-3,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
3.
{\displaystyle f(x)=3.}
When
x
=
−
1
,
{\displaystyle x=-1,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
−
1.5.
{\displaystyle f(x)=-1.5.}
When
x
=
2
,
{\displaystyle x=2,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
3.
{\displaystyle f(x)=3.}
Point
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is local maximum. Point
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is local minimum.
In the example to the right,
y
=
f
(
x
)
=
2
x
3
+
3
x
2
−
12
x
−
8
8
{\displaystyle y=f(x)={\frac {2x^{3}+3x^{2}-12x-8}{8}}}
and
y
′
=
f
′
(
x
)
=
6
x
2
+
6
x
−
12
8
.
{\displaystyle y'=f'(x)={\frac {6x^{2}+6x-12}{8}}.}
Of special interest are the points at which
f
′
(
x
)
{\displaystyle f'(x)}
or slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
f
′
(
x
)
=
0
,
x
=
−
2
{\displaystyle f'(x)=0,\ x=-2}
and
f
(
x
)
=
1.5
{\displaystyle f(x)=1.5}
or
When
f
′
(
x
)
=
0
,
x
=
1
{\displaystyle f'(x)=0,\ x=1}
and
f
(
x
)
=
−
1
7
8
.
{\displaystyle f(x)=-1{\frac {7}{8}}.}
The points
(
−
2
,
1.5
)
,
(
1
,
−
1
7
8
)
{\displaystyle (-2,1.5),\ (1,-1{\frac {7}{8}})}
are critical or stationary points of
f
(
x
)
.
{\displaystyle f(x).}
Because
y
′
{\displaystyle y'}
has exactly two real solutions for
y
′
=
0
,
f
(
x
)
{\displaystyle y'=0,\ f(x)}
has exactly two critical points.
Slope of
f
(
x
)
{\displaystyle f(x)}
to the left of
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is positive and
adjacent slope of
f
(
x
)
{\displaystyle f(x)}
to the right of
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is negative.
Therefore point
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is local maximum. Point
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is not absolute maximum.
Adjacent slope of
f
(
x
)
{\displaystyle f(x)}
to the left of
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is negative and
slope of
f
(
x
)
{\displaystyle f(x)}
to the right of
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is positive.
Therefore point
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is local minimum. Point
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is not absolute minimum.
Electric water heater
edit
Figure 1(c): Plan of county road between Town A and Town B to be constructed so that cost is minimum.
Town B is 40 miles East and 50 miles North of Town A. The county is going to construct a road from Town A to Town B.
Adjacent to Town A the cost to build a road is $500k/mile.
Adjacent to Town B the cost to build a road is $200k/mile.
The dividing line runs East-West 30 miles North of Town A.
Calculate the position of point C so that the cost of the road from Town A to Town B is minimum.
Let point
C
=
(
x
,
30
)
.
{\displaystyle C=(x,30).}
Then distance from Town A to point
C
=
x
2
+
30
2
=
x
2
+
900
.
{\displaystyle C={\sqrt {x^{2}+30^{2}}}={\sqrt {x^{2}+900}}.}
Distance from Town B to point
C
=
(
40
−
x
)
2
+
20
2
{\displaystyle C={\sqrt {(40-x)^{2}+20^{2}}}}
=
1600
−
80
x
+
x
2
+
400
{\displaystyle ={\sqrt {1600-80x+x^{2}+400}}}
=
2000
−
80
x
+
x
2
.
{\displaystyle ={\sqrt {2000-80x+x^{2}}}.}
x
=
10.52322823517
…
{\displaystyle x=10.52322823517\dots }
Figure 1(e): Sheet of cardboard to be cut and folded to make box of maximum possible volume. Cut on purple lines, fold on red lines. Design of box includes top.
A piece of cardboard of length
4
f
t
{\displaystyle 4\ ft}
and width
3
f
t
{\displaystyle 3\ ft}
will be used to make a box with a top. Some waste will be cut out of the piece of cardboard
and the remaining cardboard will be folded to make a box so that the volume of the box is maximum.
What is the height of the box?
Solving ellipse at origin
edit
Figure 1: Graph of ellipse showing semi-chord through center. When length of
t
{\displaystyle t}
is maximum, length of major axis
=
2
t
.
{\displaystyle =2t.}
When length of
t
{\displaystyle t}
is minimum, length of minor axis
=
2
t
.
{\displaystyle =2t.}
An ellipse with center at origin has equation:
A
x
2
+
B
x
y
+
C
y
2
+
F
=
0
…
(
1
)
{\displaystyle Ax^{2}+Bxy+Cy^{2}+F=0\ \dots \ (1)}
Given values
A
,
B
,
C
,
F
{\displaystyle A,B,C,F}
calculate:
In Figure 1
t
{\displaystyle t}
is any line from origin to ellipse and
θ
{\displaystyle \theta }
is angle between
X
{\displaystyle X}
axis and
t
.
{\displaystyle t.}
Aim of this section is to calculate
θ
{\displaystyle \theta }
so that length of
t
{\displaystyle t}
is maximum,
in which case length of major axis =
2
t
.
{\displaystyle 2t.}
Let
c
=
cos
(
θ
)
{\displaystyle c=\cos(\theta )}
and
s
=
sin
(
θ
)
.
{\displaystyle s=\sin(\theta ).}
Then
x
=
t
cos
(
θ
)
=
t
c
,
{\displaystyle x=t\cos(\theta )=tc,}
and
y
=
t
sin
(
θ
)
=
t
s
.
{\displaystyle y=t\sin(\theta )=ts.}
Substitute these values in
(
1
)
:
{\displaystyle (1):}
A
t
t
c
c
+
B
t
t
c
s
+
C
t
t
s
s
+
F
=
0
…
(
2
)
{\displaystyle Attcc+Bttcs+Cttss+F=0\ \dots \ (2)}
Calculate
t
′
=
d
t
d
θ
{\displaystyle t'={\frac {dt}{d\theta }}}
A
(
t
t
(
−
2
c
s
)
+
c
c
2
t
t
′
)
+
B
(
t
t
(
c
c
−
s
s
)
+
c
s
2
t
t
′
)
+
C
(
t
t
(
2
s
c
)
+
s
s
2
t
t
′
)
=
0
{\displaystyle A(tt(-2cs)+cc2tt')+B(tt(cc-ss)+cs2tt')+C(tt(2sc)+ss2tt')=0}
A
t
t
(
−
2
c
s
)
+
A
c
c
2
t
t
′
+
B
t
t
(
c
c
−
s
s
)
+
B
c
s
2
t
t
′
+
C
t
t
(
2
s
c
)
+
C
s
s
2
t
t
′
=
0
{\displaystyle Att(-2cs)+Acc2tt'+Btt(cc-ss)+Bcs2tt'+Ctt(2sc)+Css2tt'=0}
−
2
A
t
t
c
s
+
A
c
c
2
t
t
′
+
B
t
t
c
c
−
B
t
t
s
s
+
B
c
s
2
t
t
′
+
2
C
t
t
s
c
+
2
C
s
s
t
t
′
=
0
{\displaystyle -2Attcs+Acc2tt'+Bttcc-Bttss+Bcs2tt'+2Cttsc+2Csstt'=0}
+
A
c
c
2
t
t
′
+
2
B
c
s
t
t
′
+
2
C
s
s
t
t
′
−
2
A
t
t
c
s
+
B
t
t
c
c
−
B
t
t
s
s
+
2
C
t
t
s
c
=
0
{\displaystyle +Acc2tt'+2Bcstt'+2Csstt'-2Attcs+Bttcc-Bttss+2Cttsc=0}
+
t
′
(
A
c
c
2
t
+
2
B
c
s
t
+
2
C
s
s
t
)
=
+
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
{\displaystyle +t'(Acc2t+2Bcst+2Csst)=+2Attcs-Bttcc+Bttss-2Cttsc}
t
′
=
+
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
(
A
c
c
2
t
+
2
B
c
s
t
+
2
C
s
s
t
)
{\displaystyle t'={\frac {+2Attcs-Bttcc+Bttss-2Cttsc}{(Acc2t+2Bcst+2Csst)}}}
For maximum or minimum
t
:
{\displaystyle t:}
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
=
0
{\displaystyle 2Attcs-Bttcc+Bttss-2Cttsc=0}
2
A
c
s
−
B
c
c
+
B
s
s
−
2
C
s
c
=
0
{\displaystyle 2Acs-Bcc+Bss-2Csc=0}
2
A
c
s
−
2
C
s
c
=
B
c
c
−
B
s
s
{\displaystyle 2Acs-2Csc=Bcc-Bss}
Square both sides, substitute
1
−
s
s
{\displaystyle 1-ss}
for
c
c
{\displaystyle cc}
and result is:
a
S
2
+
b
S
+
c
=
0
…
(
3
)
{\displaystyle aS^{2}+bS+c=0\ \dots \ (3)}
where:
S
=
sin
2
(
θ
)
{\displaystyle S=\sin ^{2}(\theta )}
a
=
(
+
4
A
A
−
8
A
C
+
4
B
B
+
4
C
C
)
{\displaystyle a=(+4AA-8AC+4BB+4CC)}
b
=
−
a
{\displaystyle b=-a}
c
=
B
B
{\displaystyle c=BB}
Let equation of ellipse be:
55
x
2
−
24
x
y
+
48
y
2
−
2496
=
0
{\displaystyle 55x^{2}-24xy+48y^{2}-2496=0}
# python code
>>> A , B , C = 55 , - 24 , 48
>>> a = ( + 4 * A * A - 8 * A * C + 4 * B * B + 4 * C * C ); a
2500
>>> b = - a ; b
- 2500
>>> c = B * B ; c
576
>>>
>>> a , b , c = [ v / 4 for v in ( a , b , c )] ; a , b , c
( 625.0 , - 625.0 , 144.0 )
>>> S = .36
>>> a * S * S + b * S + c
0.0
>>> S = .64
>>> a * S * S + b * S + c
0.0
>>>
The solutions of quadratic equation
(
3
)
{\displaystyle (3)}
are
.36
{\displaystyle .36}
or
.64
{\displaystyle .64}
.
Therefore
sin
(
θ
)
=
±
0.6
{\displaystyle \sin(\theta )=\pm 0.6}
or
sin
(
θ
)
=
±
0.8
{\displaystyle \sin(\theta )=\pm 0.8}
.
From
(
2
)
{\displaystyle (2)}
above:
t
=
−
F
A
c
c
+
B
c
s
+
C
s
s
{\displaystyle t={\sqrt {\frac {-F}{Acc+Bcs+Css}}}}
# python code
A , B , C , F = 55 , - 24 , 48 , - 2496
t1 = ( 0.6 , 0.8 )
dict1 = dict ()
for v1 in ( t1 , t1 [:: - 1 ]) :
c1 , s1 = v1
for c in ( c1 , - c1 ) :
for s in ( s1 , - s1 ) :
t = ( - F / ( A * c * c + B * c * s + C * s * s )) ** .5
if t in dict1 : dict1 [ t ] += (( c , s ),)
else : dict1 [ t ] = (( c , s ),)
L1 = [ ( v , dict1 [ v ]) for v in sorted ([ v for v in dict1 ]) ]
for v in L1 : print ( v )
(6.244997998398398, ((0.8, -0.6), (-0.8, 0.6)))
(6.34287855135306, ((0.6, -0.8), (-0.6, 0.8)))
(7.806247497997998, ((0.8, 0.6), (-0.8, -0.6)))
(7.999999999999999, ((0.6, 0.8), (-0.6, -0.8)))
Minimum value of
t
=
6.244997998398398.
{\displaystyle t=6.244997998398398.}
Length of minor axis
=
2
∗
6.244997998398398
{\displaystyle =2*6.244997998398398}
Maximum value of
t
=
7.999999999999999.
{\displaystyle t=7.999999999999999.}
Length of major axis
=
8
∗
2
{\displaystyle =8*2}
Figure 3: Curves and values associated with car jack. When
x
=
1
,
d
y
d
x
=
0.1
…
{\displaystyle x=1,\ {\frac {dy}{dx}}=0.1\dots }
When
θ
=
45
∘
,
d
y
d
x
=
1
{\displaystyle \theta =45^{\circ },\ {\frac {dy}{dx}}=1}
When
x
=
9
,
d
y
d
x
=
2.06
…
{\displaystyle x=9,\ {\frac {dy}{dx}}=2.06\dots }
At what rate is point
A
{\displaystyle A}
moving upwards:
(a) when
x
=
9
{\displaystyle x=9}
?
(b) when
x
=
1
{\displaystyle x=1}
?
(c) when
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
?
We have to calculate
d
y
d
t
{\displaystyle {\frac {dy}{dt}}}
when
d
x
d
t
{\displaystyle {\frac {dx}{dt}}}
is given.
d
y
d
t
=
d
y
d
x
⋅
d
x
d
t
{\displaystyle {\frac {dy}{dt}}={\frac {dy}{dx}}\cdot {\frac {dx}{dt}}}
x
2
+
y
2
=
10
2
{\displaystyle x^{2}+y^{2}=10^{2}}
(equation of circle)
y
=
100
−
x
2
{\displaystyle y={\sqrt {100-x^{2}}}}
y
2
=
100
−
x
2
{\displaystyle y^{2}=100-x^{2}}
2
y
⋅
d
y
d
x
=
−
2
x
{\displaystyle 2y\cdot {\frac {dy}{dx}}=-2x}
d
y
d
x
=
−
2
x
2
y
=
−
x
100
−
x
2
{\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}={\frac {-x}{\sqrt {100-x^{2}}}}}
For convenience we'll use the negative value of the square root and say that
d
y
d
x
=
x
100
−
x
2
.
{\displaystyle {\frac {dy}{dx}}={\frac {x}{\sqrt {100-x^{2}}}}.}
Relative to line
B
C
:
{\displaystyle BC:}
When
x
=
9
,
d
y
d
x
=
2.06
,
d
y
d
t
=
2.06
⋅
5
=
10.3
{\displaystyle x=9,\ {\frac {dy}{dx}}=2.06,\ {\frac {dy}{dt}}=2.06\cdot 5=10.3}
inches
/
{\displaystyle /}
minute.
When
x
=
1
,
d
y
d
x
=
0.1
,
d
y
d
t
=
0.1
⋅
5
=
0.5
{\displaystyle x=1,\ {\frac {dy}{dx}}=0.1,\ {\frac {dy}{dt}}=0.1\cdot 5=0.5}
inches
/
{\displaystyle /}
minute.
When
θ
=
45
∘
,
x
=
10
⋅
cos
45
∘
=
7.071
{\displaystyle \theta =45^{\circ },x=10\cdot \cos 45^{\circ }=7.071}
and
d
y
d
t
=
1
⋅
5
=
5
{\displaystyle {\frac {dy}{dt}}=1\cdot 5=5}
inches
/
{\displaystyle /}
minute.
This example highlights the mechanical advantage of this simple but effective tool. When the top of the jack is low, it moves quickly.
As the jack takes more and more weight, the top of the jack moves more slowly.
Figure 4: Graph of
a
{\displaystyle a}
and